Question: Balance the following chemical equation: $ $ $\text{NH}_4\text{NO}_3 \rightarrow$ $\text{N}_2 +$ $\text{O}_2 +$ $\text{H}_2\text{O}$
Explanation: Start with the compound that has the most elements $(\text{NH}_4\text{NO}_3)$ There are $2 \text{N}$ on the left and $2 \text{N}$ on the right, so $\text{N}$ is already balanced. There are $4 \text{ H}$ on the left and $2$ on the right, so multiply $\text{H}_2\text{O}$ by ${2}$ $ \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2 + \text{O}_2 + {2}\text{H}_2\text{O} $ That gives us $3 \text{ O}$ on the left and $4$ on the right. If we try giving $\text{O}_2$ a coefficient of ${\frac{1}{2}}$ , it gives us $3 \text{ O}$ on both sides. $ \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2 + {\frac{1}{2}}\text{O}_2 + 2\text{H}_2\text{O} $ Since fractions are not usually used as coefficients, multiply everything by $2$ to get rid of the fraction. $ 2\text{NH}_4\text{NO}_3 \rightarrow 2\text{N}_2 + 1\text{O}_2 + 4\text{H}_2\text{O} $ The balanced equation is: $ 2\text{NH}_4\text{NO}_3 \rightarrow 2\text{N}_2 + \text{O}_2 + 4\text{H}_2\text{O} $